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Unformatted text preview: ME 563 – Fall 2011
Exam No. 1 Grade distribution and solution
Exam scores %
93.3
93.3
93.3
92.2
92.2
91.1
90.0
90.0
88.9
87.8
86.7
86.7
83.3
82.2
81.1
81.1
81.1
81.1
80.0
80.0
78.9
77.8
72.2
70.0
67.8
67.8
62.2
61.1
61.1
58.9
57.8
56.7
56.7
56.7
56.7
25.6 !
! Problem No. 1 (10 points)
Let [A] = [aij ] represent the ﬂexibility matrix for the threeDOF system below corresponding to
generalized coordinates q1 = x, q2 = y and q3 = θ. Determine the following components of [A]:
a22 , a12 , a31 and a23 . SOLUTION
Apply unit force at 1:
a11 = 1/k . Also, A moves same distance as O and angle of bar remains zero.
a21 = a11 = 1/k = a12 and a31 = 0.
Apply unit force at 2:
a22 = 1/k + 1/2k = 3/2k . Also, angle of bar remains zero. Therefore a32 = 0 = a23 . Therefore, Problem No. 2 (10 points)
The motion of a sixDOF system is described by the translational coordinates xj (j = 1, 2, ..., 6) as
shown below. The original conﬁguration of the system has mass and stiﬀness matrices of [M ] and
[K ], respectively. However, we do not know any other details on the system
The system is changed by: adding a spring of stiﬀness k between points 2 and 5, adding a spring
of stiﬀness k between points 3 and 6, and the addition of a point mass m at point 5, all as shown
in the bottom ﬁgure.
If the mass and stiﬀness of the modiﬁed system are written as [M ]new = [M ] + [∆M ] and
[K ]new = [K ] + [∆K ], determine the nonzero elements of the matrices [∆M ] and [∆K ]. Fill
in these nonzero elements in the matrices provided on the next page. Provide details of your
derivation in order to receive full credit for this problem.
SOLUTION
1
1
1
1
∆T = m (x2 + x5 )2 = mx2 + (2m)x2 x5 + mx2
˙
˙
˙2
˙˙
˙5
2
2
2
2
1
1
1
= M22 x2 + (M25 + M52 )x2 x5 + M55 x2
˙2
˙˙
˙5
2
2
2
Therefore, ∆M22 = ∆M25 = ∆M52 = ∆M55 = m.
1
1
1
1
1
1
∆U = kx2 + k (x6 − x3 )2 = kx2 + kx2 − (2k )x3 x6 + kx2
5
5
6
2
2
2
2
2
23
Therefore, using∆Kij = ∂ 2 ∆U/∂ xi ∂ xj gives
∆K33 = ∆K55 = ∆K66 = k , and ∆K36 = ∆K63 = −k . Problem No. 3 (10 points)
A force F acts to the right on block B of the system shown below, where x1 is the displacement of
block A and x2 is the displacement of the center of the disk O.
Determine the generalized forces Qx1 and Qx2 corresponding to F . SOLUTION
Let θ be the rotation of disk (positive in CW sense). Therefore, θ = (x2 − x1 )/R.
Using this,
B = (x2 + Rθ) ˆ = (2x2 − x1 ) ˆ
r
i
i
r
dW = F · dB = F ˆ · (2dx2 − dx1 ) ˆ = −F dx1 + 2F dx2
i
i
Therefore, Qx1 = −F and Qx2 = 2F . Problem No. 4 (10 points)
A threeDOF system, whose EOMs were derived using Lagrange’s equations, has a mass matrix of: 100
[M ] = 0 1 0 002
The ﬁrst two modal vectors are known: 1
1
X (1) = 3 X (2) = 3 1
−5 (3)
Determine the third modal vector X (3) . Scale your answer such that X1 . SOLUTION
Let X (3) 1
= a . Using orthogonality of modal vectors through [M ]:
b
0 = X (1)T [M ]X (3) T 1
100
1
= 3 0 1 0 a = 1 + 3 a + 2b
1
002
b T 1
100
1
0 = X (2)T [M ]X (3) = 3 0 1 0 a = 1 + 3a − 10b
−5
002
b 1
Solving the above: a = −1/3 and b = 0. Therefore, X (3) = −1/3 0 Problem No. 5 (10 points)
An undamped threeDOF system has mass and stiﬀness matrices of: 200
2 −1
0
2 −2 [M ] = 0 1 0 [K ] = −1
001
0 −2
1
respectively. Is the free response of the system purely oscillatory? Explain.
SOLUTION
K11 = 2 > 0
2 −1
=4−1=3>0
−1
2
2 −1
0
2 −2
−1 −2
−1
− (−1)
= −4 − 1 = −5 < 0
2 −2 = 2
−2
0
1
1
0 −2
1
From this we see that [K ] is not positive deﬁnite. Therefore, some of the eigenvalues will be real,
and as a result, the system will some nonoscillatory modal components. Problem No. 6 (10 points)
The model for an undamped threeDOF system has the following set of EOMs:
¨
[M ] + [K ] =
x
x0
This system has natural frequencies of: ω1 = 1rad/sec, ω2 = 3rad/sec and ω3 = 5rad/sec.
Damping is now added to the model with the following damping matrix:
[C ] = 0.06[M ] + 0.04[K ] (kg/sec) Determine the three modal damping ratios for the system.
SOLUTION
1α
1 0.06
0.03
ζj =
+ βωj =
+ 0.04ωj =
+ 0.02ωj
2 ωj
2 ωj
ωj
Therefore,
ζ1 = 0.03
+ 0.02(1) = 0.05
1 ζ2 = 0.03
+ 0.02(3) = 0.07
3 ζ3 = 0.03
+ 0.02(5) = 0.106
5 Problem No. 7 (10 points)
An undamped twoDOF system has a mass matrix of:
20
[M ] =
01
modal vectors of:
1
(1)
X=
2 (2) X = 1
−1 and natural frequencies of ω1 = 0 and ω2 = √ 3rad/sec. Determine the free response of the system corresponding to initial conditions of x1 (0) = 2, x2 (0) =
−2, x1 (0) = 1 and x2 (0) = 2.
˙
˙
SOLUTION
(t) = (c1 + s1 t)X (1) + (c2 cosω2 t + s2 sinω2 t)X (2)
x
˙
Note that the ICs are such that (0) = 2X (2) (⇒ c1 = 0), and (0) = X (1) (⇒ s2 = 0), and:
x
x
c2 =
X (2)T [M ] (0)
x
X (2)T [M ]X (2)
=2
=2
X (2)T [M ]X (2)
X (2)T [M ]X (2) s1 = ˙
X (1)T [M ] (0)
x
X (1)T [M ]X (1)
=
=1
X (1)T [M ]X (1)
X (1)T [M ]X (1) Therefore,
( t) =
x 1
2 t+2 1
−1 √
cos 3t Problem No. 8 (10 points)
The characteristic equation for a rod problem has been found to be:
2
cotβ L = −1
βL
ρ
where β = E ω . • Sketch the components of the characteristic function in the plot provided below.
• Place upper and lower bounds on all natural frequencies for the system. SOLUTION 2j − 1
π
2 E
< ωj < j π
ρL2 E
ρL2 Problem No. 9 (10 points)
An undamped twoDOF system with harmonic excitation has EOMs of:
21
x1
¨
4 −1
x1
1
+
=
sinΩt
12
x2
¨
−1
1
x2
1
Determine all frequencies Ω corresponding to antiresonances in the particular solution P (t). I
x
SOLUTION
[H ] = −2Ω2 + 4 −Ω2 − 1
−Ω2 − 1 −2Ω2 + 1 −1 1
=
∆ −2Ω2 + 1
Ω2 + 1
2+1
Ω
−2Ω2 + 4 where ∆ = det[H ]. Therefore,
2
1
Ω2 + 1
1
− Ω2 + 2
= 1 −2Ω + 1
X
=
Ω2 + 1
−2Ω2 + 4
1
∆
∆ − Ω2 + 5
From this we have:
X1 = − Ω2 + 2
=0
∆ when Ω= √ 2 X2 = − Ω2 + 5
=0
∆ when Ω= √ 5 ...
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This document was uploaded on 12/23/2011.
 Fall '09

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